Therefore, if the equation of the line can be found then the e.m.f. can also be found. By using the results from the gradient calculations I know that the gradient is -0.47 � 0.03. If I use a pair of co-ordinates that lie on the line I can find the constant value, C, using the equation y - y1 = m (x - x1). Using co-ordinates (0.2,1.415) y - 1.415 = -0.47 (x - 0.2) y - 1.415 = -0.47x + 0.094 y = -0.47x + 1.509 This means that C must equal 1.51 (3sf). To find the error in this value I substituted the gradients and a set of points for the minimum line and maximum line in to the equation E = Ir + V. For the maxm line: E = 0.86 x 0.49 + 1.1 E = 1.52 (3sf) For the minm line: E = 1.0 x 0.45 + 10.5 E = 1.50 (3sf) The difference between these two values and the values calculated is 0.01. Therefore the electromotive force of the cell = 1.51�0.01V(3sf) Evaluating Evidence These were the possible sources of error: * As I was using an analogue ammeter the needle that is used to read the values off is slightly higher than the actual scale so to enable a completely accurate reading you have to look directly down at it. ...read more.

Conclusion

Although the ammeter could have been more accurate by having a smaller error and it may even be more precise if two or even three people were to perform the experiment together. One to turn the circuit on and off, one to take the voltmeter reading and one to take the ammeter reading. This way there is no loss of accuracy due to the time delay between taking readings that is caused by the operator. As the technique used was very accurate and precise, this made the conclusions drawn also very accurate. Although there may have been a human error made when drawing the best-fit, maximum and minimum lines on the graph and reading the results off accurately. To confirm my final value of e.m.f. I found the e.m.f. of the cell in the circuit each time I took a reading. These were the results. Current (Amps) E.M.F. taken from Voltmeter 1 2 3 Average 0.1 1.50 1.51 1.51 1.51 0.2 1.51 1.51 1.50 1.51 0.3 1.51 1.52 1.50 1.51 0.4 1.50 1.51 1.50 1.50 0.5 1.52 1.50 1.50 1.51 0.6 1.51 1.50 1.50 1.50 0.7 1.50 1.50 1.50 1.50 0.8 1.49 1.50 1.49 1.49 0.9 1.50 1.49 1.48 1.49 1.0 1.48 1.49 1.48 1.48 These show that the e.m.f value is actually around 1.50V, compared with my final result of 1.51V. This shows that the techniques appear to be very accurate and precise. This table also shows that at times the value overlaps the result I found which adds weight to the conclusion that my answer is accurate. Alice Salt ...read more.

During the investigation I will measure, the voltmeter readings, the ammeter readings which will be in milliamps because these are appropriate units to use. These will be approximately to three s.f. or to the nearest thousandth, because calculating to the nearest amp will result in all readings being near enough

line of best fit with the biggest difference to the normal line of best fit to calculate the Emf + the error. Emf of the maximum line of best fit = 1.75v Emf of the minimum line of best fit = 1.66v Hence, the maximum line of best fit has the bigger difference.

Secondly, the rheostat should be set to its maximum value in the beginning of the experiment, so that the current is the lowest at first, then increase gradually. As high current produces heat, it would increase the resistance of the connecting wires and the internal resistance.

product moment correlation coefficient reflects the degree of linear relationship between two variables. It ranges from +1 to -1. A correlation of +1 means that there is a perfect positive linear relationship between variables. A correlation of -1 means that there is a perfect negative linear relationship between variables.

I will also compare where the curves peak; at what distance does the curve peak (earliest/latest) in relation to the load resistance. Method: The following method was used to gain the data I need to process. Knowing how the data was collected will help me ascertain where errors may have occurred and how they may have been prevented.

The exception being if it were, for example, 0.20A it would simply be recorded as 0.2A. Another problem is that with the voltage the values are constantly changing. To overcome and improve the situation the highest value seen on the voltmeter added to the lowest value seen all divided by two should give the average voltage for that particular repeat.

� 30.4 2 170.0 170.2 169.0 3 176.0 190.0 178.2 Modifications The results we obtained were non-precise and irregular, so the graph plotted is not linear. From the graph before it is obvious that there is little correlation and the error uncertainty associated with the results is very large.